Gh plane the number poles of the loop transfer

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Unformatted text preview: M 381 - Stability 18 The Nyquist Criterion: A Simple Example • The number of encirclements of the –1 point in The point the GH-plane is zero, N = 0. GH plane • The number poles of the loop transfer function The GH(s) in the RHS, is zero (P = 0). Note: the GH ). Nyquist path excludes the pole at the origin. Nyquist • Therefore, the number of poles of the closed-loop loop system = the zeroes of 1 + GH(s), in the RHS, is GH ), Z=N+P =0+0 =0. • The closed-loop system is stable. 12/13/2008 HSK - EPM 381 - Stability 19 The Nyquist Criterion: Sketching the GH(s) Function • In most cases only an approximate sketch of the In GH(s) mapping is required. GH • Consider various segments of the Nyquist path. jω j∞ II × III σ I IV -j∞ 12/13/2008 Section I : s = ρ e jθ ; ρ → 0 θ = –90 → +90 Section II : s = j0+ → j∞ Section III : s = R e jθ ; R → ∞ θ = +90 → –90 +90 Section IV : s = – j∞ → j0– HSK - EPM 381 - Stability 20 The Nyquist Criterion: Sketching the GH(s) Function GH • Consider a general loop transfer function K( s + z1 )( s + z2 )( )L GH ( s ) = k s ( s + p1 )( s + p2 )( )L • Section I: ( s = ρ e jθ , ρ → 0 , θ = – 90º → + 90º ) θ K ( ρ e j + z1 )( L ) θ GH ( ρ e j ) = k j kθ ρ e ( ρ e jθ + p1 )( L ) As θ varies from as ρ → 0 – 90º→ +90º (CCW) Kz1z 2 L /GH varies from varies → ∞ / − kθ GH → k j kθ ρ e p1p2 L k 90º → – k 90º (CW) 90 90 12/13/2008 HSK - EPM 381 - Stability 21 The Nyquist Criterion: Sketching the GH(s) Function GH • Section II: (s = jω , ω = 0+→ ∞) ω → 0+ ω →∞ ° GH ( jω ) → ∞ / − k 90...
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